Question: Let $a \bowtie b = a+\sqrt{b+\sqrt{b+\sqrt{b+...}}}$. If $4\bowtie y = 10$, find the value of $y$.
Solution: We know that $$4\bowtie y = 4+\sqrt{y+\sqrt{y+\sqrt{y+...}}}=10.$$Therefore, $\sqrt{y+\sqrt{y+\sqrt{y+...}}}=6$. Because the series of $\sqrt{y+\sqrt{y+\sqrt{y+...}}}$ is infinite, we can substitute $6$ into the series for any $\sqrt{y+\sqrt{y+\sqrt{y+...}}}$ we want. Thus, $$\sqrt{y+\sqrt{y+\sqrt{y+...}}}=6$$implies that $$\sqrt{y+\sqrt{y+\sqrt{y+...}}}=\sqrt{y+6}=6.$$Squaring both sides of this new equality, we have $y+6=36$, or $y=\boxed{30}$.